Physics 0625 · syllabus 1.6
Cambridge IGCSE Physics 0625 — Momentum topical past papers
Momentum and impulse, conservation of momentum in collisions and explosions, and force as the rate of change of momentum: Extended content, set on Papers 2 and 4.
- Questions
- 511
- Years
- 2016–2026
- With mark scheme
- 100%
- Questions 2022–2026
- 230
Anyone can browse these pages. Opening the questions and their mark schemes takes a free Momentum account.
What Momentum covers
Momentum is mass multiplied by velocity, p = mv, measured in kg m/s. Because velocity has a direction, so does momentum: choose one direction as positive, and give anything moving the other way a negative velocity. The whole topic is Supplement content, so it is examined only on the Extended papers, Papers 2 and 4.
A force acting for a time changes an object's momentum. The product of the force and the time, FΔt, is the impulse, measured in N s, and it equals the change in momentum, Δ(mv). Put the other way round, the resultant force equals the rate at which momentum changes, F = Δp/Δt. That is why crumple zones, airbags, cycle helmets and bending your knees as you land all protect you: they spread the same change in momentum over a longer time, so the force is smaller.
When objects collide or push apart and no external resultant force acts on them, the total momentum before equals the total momentum after. You apply this in one dimension: trolleys that collide and join, a ball that rebounds, a gun that recoils, two skaters who push off from rest and glide apart with equal and opposite momenta. Kinetic energy is usually not conserved in these events, because some is transferred to internal energy and sound, but the total momentum is.
In the syllabus
The headings this topic has in the Cambridge International 2026–2028 syllabus, as the board prints them.
- 1.6 Momentum
What you should be able to do
- Calculate momentum with p = mv, using a sign to show direction
- Use impulse = FΔt = Δ(mv) to find a force, a contact time or a change in velocity
- Apply conservation of momentum to collisions, objects that join and explosions in one dimension
- Use F = Δp/Δt and explain why a longer impact time means a smaller force
- Explain safety features such as crumple zones, airbags and helmets in terms of momentum and time
Key ideas
- Momentum p = mv, a vector, in kg m/s
- Impulse = FΔt = Δ(mv), in N s; 1 N s = 1 kg m/s
- Resultant force F = Δp/Δt
- Conservation of momentum: total momentum before = total momentum after, when no external resultant force acts
How Momentum has been examined
From 2022 to 2026, Momentum was on Paper 4 · Theory (Extended) in 14 of 14 exam series, worth about 18% of that paper’s marks.
| Paper | Questions | Share of marks | Series |
|---|---|---|---|
| Paper 2 · Multiple Choice (Extended) | 172 | 13% | 14 of 14 |
| Paper 4 · Theory (Extended) | 58 | 18% | 14 of 14 |
Momentum’s own count, over the questions in its bank for these years: a series is one exam session, every zone included, and a question filed under two chapters counts in both. Median marks per question: 1. The bank files this topic together with neighbouring ones, so these figures count every question in its chapters, including some on those topics.
Filed in Momentum under: Forces and Motion.
A worked example
A trolley of mass 1.5 kg moving at 4.0 m/s collides with a stationary trolley of mass 2.5 kg, and they stick together. Find their velocity just after the collision. The collision lasts 0.050 s; find the average force on the 2.5 kg trolley.
- Total momentum before = 1.5 × 4.0 + 2.5 × 0 = 6.0 kg m/s
- Afterwards the joined trolleys have a mass of 4.0 kg: 4.0 × v = 6.0, so v = 1.5 m/s in the original direction
- Change in momentum of the 2.5 kg trolley = 2.5 × 1.5 − 0 = 3.75 kg m/s
- Average force = Δp/Δt = 3.75 / 0.050 = 75 N
- Check: the 1.5 kg trolley loses 6.0 − 1.5 × 1.5 = 3.75 kg m/s, equal and opposite to the other trolley's gain
Answer: 1.5 m/s in the original direction; an average force of 75 N
Written by Momentum for this page, with its own numbers; it is not taken from a past paper.
Mistakes that cost marks
- Ignoring direction when an object rebounds — a 0.20 kg ball that hits a wall at 5.0 m/s and bounces back at 3.0 m/s changes its momentum by 0.20 × 8.0 = 1.6 kg m/s, not 0.40 kg m/s
- Forgetting that the total momentum is zero before an explosion from rest — afterwards the pieces have equal momenta in opposite directions, so the total is still zero
- Giving two objects one shared velocity after a collision in which they move apart — only objects that join move off together; otherwise each needs its own velocity term
- Using kinetic energy to find a velocity after a collision — kinetic energy is usually not conserved, because some becomes internal energy and sound, whereas momentum is
- Writing the unit of momentum as kg/m s or N/s — it is kg m/s, which is the same as N s
Exam technique
- Draw a before-and-after sketch with every mass, every velocity and the positive direction marked
- Write the conservation statement in words — total momentum before = total momentum after — and then substitute
- For an average force, find the change in momentum first, then divide by the contact time in seconds, not milliseconds
- In safety explanations, link a longer time for the same change in momentum to a smaller force, and quote F = Δp/Δt
Practising in Momentum
- Every question in a chapter as one list, from every paper and year in the bank, newest first or shuffled.
- Each question turns over to its own part of the mark scheme, where the board published one.
- Narrow by paper, year, session, zone and marks, and star the questions worth keeping.
- Pick chapters, a number of questions and a time, and the exam builder draws a mock to sit and mark.
Past-paper questions and mark schemes are the copyright of Cambridge International. Momentum shows them to signed-in learners for revision and does not publish them on these pages.